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So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .

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  • $\begingroup$ How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me. $\endgroup$ – amd Jan 12 at 9:32
  • $\begingroup$ Why do we need calculus to begin with? Analytic geometry works perfectly well. $\endgroup$ – Oscar Lanzi Jan 12 at 11:11
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Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...

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  • $\begingroup$ The OP asked for a solution that uses calculus. $\endgroup$ – amd Jan 12 at 9:23
  • $\begingroup$ No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful. $\endgroup$ – Oscar Lanzi Jan 12 at 11:15
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We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:

$$ (x-h) = a \cos t,\, (y-k)= a \sin t; \tag1 $$ and solve for the parameter

$$ \frac{y-y_1}{x-x_1}= \frac{k+ a \sin t -y_1}{h+ a \cos t -x_1}=\frac{-\cos t} {\sin t}\tag2$$

the latter has been obtained by chain rule differentiation

Simplifying get the condition to obtain $t_1$ of tangent points (two)

$$ \sin t_1 (k-y_1)+ \cos t_1 (h-x_1)+a=0 \tag3 $$

It can be also recognized as tangent-normal form of a straight line.

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Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(R\cos\theta, R\sin\theta)$.
Let $E = (a, b)$ be the external point.
Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $\vec{CM}$ is orthogonal to $\vec{EM}$.

It immediately gives: $$\cos\theta\,(x_0+ R\cos\theta - a)+ \sin\theta\,(y_0+R\sin\theta-b) = 0$$ And then $$\cos\theta\,(x_0-a) + \sin\theta\,(y_0-b) + R$$

One possibility to go further is to rewrite $\vec{EC}$ as $(\rho\cos\phi, \rho\sin\phi )$
And therefore $$\cos(\theta - \phi) = -\frac{R}{\rho}$$

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